Phosphorus and chlorine gases combine to produce phosphorus trichloride: (g) + (g) → (g) Δ_rG° at 298 K for this reaction is -642.9 kJ mol^-1. The value of Δ_rG at 298 K for a reaction mixture that consists of and is ________ kJ mol^-1.

Question

Quizplus · Accepted Answer

The change in Gibbs free energy (ΔG) of a reaction is related to its standard free energy change (ΔG°) and the reaction quotient (Q) by the equation: ΔG = ΔG° + RT ln(Q), where R is the gas constant and T is the temperature in Kelvin. At constant temperature and pressure, the change in Gibbs free energy is related to the change in entropy (ΔS) by the equation:
ΔG = ΔH - TΔS, where ΔH is the change in enthalpy.

In this problem, we are given the value of ΔS° for the reaction and asked to find ΔG at a certain temperature for a reaction mixture with known concentrations of reactants and products. To find ΔG, we need to first find Q, which is the reaction quotient. For a gas-phase reaction, Q is the product of the partial pressures of the products raised to their stoichiometric coefficients divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients. Thus, we have:
Q = (P(PCl3))^11 / (P(P4) * P(Cl2))^11
where P is the partial pressure of each gas in the mixture.

We are not given the partial pressures of the gases in the mixture, so we cannot calculate Q directly. However, we can use the fact that the reaction is at equilibrium to relate Q to the equilibrium constant K:
K = Q at equilibrium
At equilibrium, the forward and reverse reactions have equal rates, so there is no net change in the concentrations of reactants and products. Therefore, we can assume that the concentrations of the reactants and products in the mixture are at equilibrium and use K to find Q:
K = (P(PCl3))^11 / (P(P4) * P(Cl2))^11
-642.9 kJ/mol = -RT ln(K)
ln(K) = 642.9 kJ/mol / (8.314 J/mol-K * 298 K) = -81.3
K = e^(-81.3) = 7.79 × 10^(-36)

Now we can use the expression for ΔG in terms of ΔG° and Q to find ΔG for the reaction mixture:
ΔG = ΔG° + RT ln(Q)
= -642.9 kJ/mol + 8.314 J/mol-K * 298 K * ln[(P(PCl3))^11 / (P(P4) * P(Cl2))^11]
= -642.9 kJ/mol + 2.48 J/mol * ln[(P(PCl3))^11 / (P(P4) * P(Cl2))^11]

To find the value of ΔG for the given mixture, we need to make an additional assumption about the relative magnitudes of the partial pressures of the gases. One reasonable assumption is that the total pressure is constant and equal to 1 atm, so we can use mole fractions to relate the partial pressures to the number of moles of each gas:
P(PCl3) = x(PCl3) * P(total) = 11/(11+1+3) * 1 atm = 0.846 atm
P(P4) = x(P4) * P(total) = 1/(11+1+3) * 1 atm = 0.077 atm
P(Cl2) = x(Cl2) * P(total) = 3/(11+1+3) * 1 atm = 0.077 atm

Substituting these values into the expression for ΔG, we get:
ΔG = -642.9 kJ/mol + 2.48 J/mol * ln[(0.846 atm)^11 / (0.077 atm)^11 * (0.077 atm)^(-11)]
= -642.9 kJ/mol + 2.48 J/mol * ln(1.22 × 10^16)
= -642.9 kJ/mol + 88.8 kJ/mol
= -554.1 kJ/mol

Thus, the answer is E, -647.8 kJ/mol.

Phosphorus and Chlorine Gases Combine to Produce Phosphorus Trichloride